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Standard XII

Physics

The 'e' Equation

A plane flyin...

Question

A plane flying horizontally at a height of $1500 m$ with a velocity $200 m s^{- 1}$ passes directly overhead on antiaircraft gun. Then, the angle with the horizontal at which the gun should be fired from the shell with muzzle of $400 m s^{- 1}$ to hit the plane is

90^{\circ}

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60^{\circ}

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30^{\circ}

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45^{\circ}

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Solution

The correct option is B $60^{\circ}$
Horizontal distance covered should be same at the time $t$ of collision
$400 cos θ \times t = 200 \times t ⟹ cos θ = \frac{1}{2} o r θ = 60^{\circ}$

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A plane flying horizontally at a height of 1500 m with a velocity 200ms−1 passes directly overhead on antiaircraft gun. Then, the angle with the horizontal at which the gun should be fired from the shell with muzzle of 400 ms−1 to hit the plane is

The correct option is B 60∘Horizontal distance covered should be same at the time t of collision400cosθ×t=200×t⟹cosθ=12orθ=60∘

A plane flying horizontally at a height of $1500 m$ with a velocity $200 m s^{- 1}$ passes directly overhead on antiaircraft gun. Then, the angle with the horizontal at which the gun should be fired from the shell with muzzle of $400 m s^{- 1}$ to hit the plane is

The correct option is B $60^{\circ}$
Horizontal distance covered should be same at the time $t$ of collision
$400 cos θ \times t = 200 \times t ⟹ cos θ = \frac{1}{2} o r θ = 60^{\circ}$